heureka

z^2=i  solve z

$$\small{\text{$
\begin{array}{rcl}
z^2&=&i \qquad | \qquad z^2= a+bi \quad a=0 $ and $ b = 1\\\\
r&=& \sqrt{a^2+b^2}=\sqrt{0^2+1^2}=\sqrt{1^2}=1\\\\
\varphi &=& \arctan{ \left(\dfrac{b}{a}\right) }=\dfrac{1}{0} \Rightarrow \varphi = \dfrac{\pi}{2}
\qquad
\varphi =
\begin{cases}
\arctan\left(\frac{b}{a}\right) & \text{for}\ a > 0,\ b \geq 0 \qquad \text{(1. Quadrant)}\\
\arctan\left(\frac{b}{a}\right) + 2\pi & \mathrm{f\ddot ur}\ a > 0,\ b < 0 \qquad \text{(4. Quadrant)}\\
\arctan\left(\frac{b}{a}\right) + \pi & \mathrm{f\ddot ur}\ a < 0 \qquad\qquad\quad \text{(2./3. Quadrant)}\\
\frac{\pi}{2} & \mathrm{f\ddot ur}\ a = 0,\ b > 0 \qquad \text{(positive Im-axis)}\\
\frac{3\pi}{2} & \mathrm{f\ddot ur}\ a = 0,\ b < 0 \qquad \text{(negative Im-axis)}\\
\text{indeterminate} & \mathrm{f\ddot ur}\ a = 0,\ b = 0 \qquad \text{(zero)}
\end{cases} \\\\
z^2&=& r\cdot e^{i\cdot \varphi}\\\\
z^2&=& 1\cdot e^{i\cdot \frac{\pi}{2}+2k\pi} \qquad | \qquad \sqrt{}\\\\
z_k &=& e^{i\cdot \frac{\pi}{2\cdot 2 }+\frac{2k\pi}{2} } \\\\
z_k &=& e^{i\cdot \frac{\pi}{4}+ k\pi } \\\\
k =0 \quad z_0 &=& e^{i\cdot \frac{\pi}{4}}\\\\
k =1 \quad z_1 &=& e^{i\cdot \frac{5\pi}{4}} \\\\
&& \boxed{~e^{i\varphi} =\cos{(\varphi)+i\sin{(\varphi)}} ~}\\\\
z_0 &=& e^{ i\frac{\pi}{4} } =\cos{(\frac{\pi}{4})+i\sin{(\frac{\pi}{4})}}\\\\
\mathbf{z_0} &\mathbf{=}& \mathbf{\dfrac{ \sqrt{2} }{2} +\dfrac{ \sqrt{2} }{2} i}\\\\
z_1 &=& e^{ i\frac{5\pi}{4} } =\cos{(\frac{5\pi}{4})+i\sin{(\frac{5\pi}{4})}}\\\\
\mathbf{z_1} &\mathbf{=}& \mathbf{-\dfrac{ \sqrt{2} }{2} -\dfrac{ \sqrt{2} }{2} i}
\end{array}
$}}$$

$$1288,80\cdot \dfrac{10}{100}\\\\
=1288,80\cdot\dfrac{1}{10}\\\\
=128,88$$

I set y are her friends and x are the sweets.

Then we have first 8y-5=x and second 6y+19=x also.

We can set equal 8y-5=6y+19

So we have 2y=24 or y=12.

She has 12 friends.

We go back to the first equation and set for y the number 12.

8*12-5=x and x=96-5 or x=91

She has 91 sweets

$$13\cdot 0,10+12\cdot x=6\\
1,3+12x=6\\
12x=4,7\\\\
x=\dfrac{4,7}{12}\\\\
x=0,3917~m$$

Der Abstand der Balken betraegt rund 39,17cm

$$c=2\pi r=29
\\\\
h=r\cdot
\tan{(57)}\\\\
h=\dfrac{29}{2\pi}\cdot\tan
{(57)}\qquad r=4.62 ~m
\\\\
h=4.61549335\cdot 1.539864964\\\\
h=7.1072365~m\\\\
h
=
7.11~m$$

$$\dfrac23\cdot\dfrac{16}{16}=\dfrac{32}{48}\\\\
\dfrac34\cdot\dfrac{12}{12}=\dfrac{36}{48}\\\\\\
\text{between }\\
\dfrac{33}{48}=\frac{11}{16}\\\\
\dfrac{34}{48}=\frac{17}{24}\\\\
\dfrac{35}{48}$$

Aus 60 Liter 96%tigem Alkohol und einer unbekannten

Menge 80%tigem Stroh Rum soll 90%iter Alkohol hergestellt werden.

Wieviel 80%igen Rum brauche ich?

 $$\small{\text{
\boxed{
$V_1 \cdot p_1 + V_2 \cdot p_2 = (V_1 + V_2)\cdot \mathbf{p}\qquad $
wobei $\mathbf{p}$ den Prozentsatz der Gesamtmischung darstellt.
}
}}\\\\
\small{\text{
$\textcolor[rgb]{150,0,0}{V_1 = 60} $ Liter und $\textcolor[rgb]{150,0,0}{p_1 = 96}\%$
}}\\
\small{\text{
$\textcolor[rgb]{0,150,0}{V_2 = x} $ Liter und $\textcolor[rgb]{0,150,0}{p_2 = 80}\%$
}}\\
\small{\text{
$\mathbf{p = 90}\%$}}\\\\
\small{\text{$
\begin{array}{rcl}
\textcolor[rgb]{150,0,0}{V_1 \cdot p_1} +\textcolor[rgb]{0,150,0}{ V_2 \cdot p_2}
&=& (\textcolor[rgb]{150,0,0}{V_1} + \textcolor[rgb]{0,150,0}{V_2} )\cdot \mathbf{p}\\
\textcolor[rgb]{150,0,0}{60 \cdot 96} +\textcolor[rgb]{0,150,0}{ x \cdot 80}
&=& (\textcolor[rgb]{150,0,0}{60} + \textcolor[rgb]{0,150,0}{x} )\cdot \mathbf{90}\\
\textcolor[rgb]{150,0,0}{60 \cdot 96} +\textcolor[rgb]{0,150,0}{ x \cdot 80}
&=& \textcolor[rgb]{150,0,0}{60}\cdot \mathbf{90} + \textcolor[rgb]{0,150,0}{x}\cdot \mathbf{90}\\
\textcolor[rgb]{150,0,0}{60 \cdot 96} - \textcolor[rgb]{150,0,0}{60}\cdot \mathbf{90}
&=& \textcolor[rgb]{0,150,0}{x}\cdot \mathbf{90}-\textcolor[rgb]{0,150,0}{ x \cdot 80} \\
\textcolor[rgb]{150,0,0}{60}\cdot 6 &=& \textcolor[rgb]{0,150,0}{x}\cdot 10 \\
\textcolor[rgb]{0,150,0}{x}\cdot 10 &=& \textcolor[rgb]{150,0,0}{60}\cdot 6 \\
\textcolor[rgb]{0,150,0}{x}\cdot 10 &=& 360 \\\\
\textcolor[rgb]{0,150,0}{x} &=& \dfrac{360}{10}\\\\
\textcolor[rgb]{0,150,0}{x} &\mathbf{=}& \mathbf{36} \\
\end{array}
$}}$$
 

Du brauchst 36 Liter 80%igen Rum.

What is the smallest number that can be divided by 5, 7 and 11, leaving a remainder of 4 in each case?

$$x= 4+z\cdot 385 \qquad z \in Z$$

 $$\small$\text{
Solve for x : $\log_9{(x^2)}=9$
}}$$

$$\small{\text{Formula: $\boxed{~ \log_b{(b^a)}=a~}$}}\\\\
\small{\text{
We have $b=9$, and $a = 9$ so $\log_9{(9^9)}=9$, but $9^9$ is $x^2$
}}\\\\
\small{\text{$
\begin{array}{rcl}
x^2 &=& 9^9\\
x &=& 9^{\frac{9}{2}}\\
x &=& \left(\sqrt{9} \right)^9\\
x &=& \left(\sqrt{3^2} \right)^9\\
x &=& 3^9\\
\mathbf{x} & \mathbf{=} & \mathbf{19683}\\
\\
\hline
\\
\end{array}
$}}\\$$